lefteris_kaliamboswikiaorg-20200214-history
STRUCTURE OF INDIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Indium (In) consists of two primordial nuclides, with the most common (~ 95.7%) nuclide (In-115) being measurably though weakly radioactive. Its spin-forbidden decay has a half life of 441 trillion years. The stable isotope In-113 is only 4.3% of naturally occurring indium. Among elements with a known stable isotope, only tellurium and rhenium similarly occur with a stable isotope in lower abundance than the long-lived radioactive isotope. Other than In-115, the longest-lived radioisotope is In-111 with a half-life of 2.8047 days. All other radioisotopes have half-lives less than a day. This element also has 47 isomers, the longest-lived being In-114m1 with a half-life of 49.51 days. All other meta-states have half-lives less than a day, most less than an hour, and many measured in milliseconds or less. ' ' STRUCTURE OF In-97, In-99, In-101, In-103, In-105, In-107, In-109, In-111, In-113, AND In-115 WITH S =+9/2 ' For understanding the structure of this group (which includes the one stable isotope, the In-113) you must read my STRUCTURE OF In-113. In the following diagram of In- 98 with S= 0 based on Cd-96 we clear that the additional p49 and n49 makes a vertical rectangle with S =0 in front of n15p17 and breaks the high symmetry of Cd-96 ' ''' '''DIAGRAM OF In-98 WITH S = 0 In this structure you see the six horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5, and -HP6 along with the horizontal squares of opposite spins like the -HSQ and +HSQ. Here the p49and n49 which make a vertical rectangle with S=0 in front of n15p17 are not shown. Note that the p45, n45, p47, n47, p46, n46, p48, and n48 make the symmetrical alpha particles of high symmetry. Also the p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins are not shown. Moreover the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12........n12......p32' ' -HP6 p31....n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' p47......n27.........p8..........n8.......p28....n48' ' -HP4 n45.....p27.......n7..........p7.......n28.........p46 ' ' n47......p25.........n6.........p6.......n26....p48' ' +HP3 p45.....n25……p5........n5……...p26.........n46 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2……p2........n22' ' +HP1 n21......p1........n1........p22 ' ' p37......n37 ' ' -HSQ n39......p39 ' Then in the presence of an odd number of extra neutrons the structure of In-98 becomes a new structure of In-98 with S = +4 because the p37n37 and p39n39 change their spins from S =-2 to S =+2 giving S = +4 Particularly they move from the -HSQ to +HSQ in order to make horizontal bonds with p38n38 and n40p40. Under this condition in the absent of one neutron of negative spin one gets the structure of In -97 with S= +9/2. That is S = +4 - 1(-1/2) = +9/2 On the other hand in the presence of the extra n50(+1/2) we get the structure of In-99 with . S =+9/2. That is S = +4 + 1(+1/2) = +9/2. Similarly in the presence of an odd number of extra neutrons giving S = +1/2 we get the structures of the unstable nuclides existing from In-101 to In-111. Note that the extra neutrons of them make two bonds per neutron but the small number cannot gives enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structure of In-113 the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. Whereas the two more extra neutrons of In -115 in the absence of blank positions make single bonds leading to the decay. STRUCTURE OF In-117, In-119, In-121, In-123, In-125, In-127, In-129, In-131, AND In-135 WITH S = +9/2 Similarly the structure of the above unstable nuclides is based on the same new structure of the In-98 with S = +4, because the odd number of extra neutrons gives S =+1/2. For example the In-135 has 37 extra neutrons which give S = +1/2. (19 neutrons of positive spins and 18 neutrons of negative spins). Note that here the more extra neutrons than those of In-115 make single bonds leading to the decay. STRUCTURE OF In-112, In-114, In-116, In-118, In-120, AND In-122 WITH S = +1 After a careful analysis I found that the structure of the above unstable nuclides having an even number of extra neutrons is based on a new structure of In-98 with S = +1, because the additional n49p49 with S =+1 makes horizontal bonds with the p38n38. Under this condition the structures of the above nuclides have extra neutrons of opposite spins giving S =0 . For example the In-122 has 24 extra neutrons of opposite spins. Note that they cannot give stable structures because the additional n49p49 with S = +1 breaks the high symmetry. 'ST RUCTURE OF In-102, In-104, In-106, In-108, AND In-110 ' After a careful analysis I found that the structure of the above unstable nuclides having an even number of extra neutrons is based on a new structure of In-98 with S = +5, because in the structure of the In-98 with S= +4 the additional p49n49 with S =+1 makes horizontal bonds with the n38p40. That is S = +4 +1 = +5. For example the In-102 with S = +6 has two extra neutrons of positive spins and two extra neutrons of opposite spins giving S =0. That is S = +5 + 2(+1/2) +0 = +6 Whereas the In -110 with S = +7 has 4 extra neutrons of positive spins and 8 extra neutrons of opposite spins giving S =0. That is S = +5 + 4(+1/2) + 0 = +7. 'STRUCTURE OF In-124, In-126, AND In-128 WITH S = +3 ' After a careful analysis I found that the structure of the above unstable nuclides having an even number of extra neutrons is based on a new structure of In-98 with S = +3, because the additional n49p49 with S =+1 makes horizontal bonds with the p38n38. and the p37n37 changes the spin from S= -1 to S =+1 giving S =+2 . That is S =+1 +2 = +3. Particularly it moves from the -HSQ to the +HSQ in order to make horizontal bonds with n40p40. Under this condition the structures of the above nuclides have extra neutrons of opposite spins giving S =0 . For example the In-128 has 30 extra neutrons of opposite spins. Category:Fundamental physics concepts